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[Emil Jeřábek posted a similar comment while I was writing this…] Probably linear programming is the simplest way: Let's say that $I_i = [l_i,u_i]$. Now plug the following linear program into any li …

I guess, uniqueness depends on the rank of $X$. I would also suggest to look at the reformulate with vectorization as $$\newcommand{\vec}{\operatorname{vec}} \|X-XC\|_F^2 = \|\vec(X) - (I\otimes X)\v …

I think the relation is not that simple and here is why: Since $A_1 = Q_1R_1$ we have $$ A_2 = \begin{bmatrix}A_1\\ I\end{bmatrix} = \begin{bmatrix}Q_1 & 0\\0 & I\end{bmatrix}\begin{bmatrix}R_1\\I\en …

I know that eigenvalue estimates involving products of matrices are in general tricky, but probably this question has some hope: Let $A$ and $B$ be two real symmetric positive semi-definite $n\times …

It makes a considerable difference if you need to project a vector just once or repeatedly inside some loop of another algorithm. If you would be in the latter case than it would be indeed a good idea …

Not an answer but too many equations for a comment: $\newcommand{\vec}{\mathrm{vec}}$ To compute an element $v\in\mathbb{R}^{3n^2}$ of the kernel of $$M = \begin{bmatrix} A\otimes I\\\\ I\otimes B\en …

Also a borderline suggestion since it is rather multilinear than just linear: Recent progress on low rank tensor approximation for all kinds of different applications within mathematics. A list of app …

If you just ask about the relation between $A = USV^T$ and $\tilde A = U\tilde SV^T$ it is just that (for the spectral or the Frobenius norm) it holds that $$ \|A-\tilde A\| = \|USV^T - U\tilde SV^T\| …

Phillip Lampe seems to be correct. Here are the eigenvalues and eigenvectors computed by hand: Let $k_1 = 2 + \tfrac12 + \cdots + \tfrac{1}{N-1}$, then: $\lambda_0 = 0$ with eigenvector all ones (by …

Depending on the matrix free method: If it is iterative, you may just initialize it with a nonzero vector $x_0$. For example the Richardson iteration $x_{k+1} = x_k - A^T Ax_k$ does converge to the pr …

I do not know a definitive "weakest" condition and I doubt there is one. Many results in the realm of Hahn-Banach do the trick, i.e. there is the general result for $A,B$ convex and $A$ open (both ope …

This is also related to equiangular tight frames (ETFs). A frame is a kind of "overcomplete basis" of an inner product space; more precisely a family $(f_1,\dots,f_n)$ of vectors of an inner product s …

I suspect that you mean $0 < t < 1$ in your question. Then the answer in still no under the added condition that both matrices are not of full rank. Consider $$ A_1 = \begin{pmatrix}1 & 0 & 0\\\\ 0 & …

If you square your equations to get $|\langle x,a_i\rangle|^2 = b_i^2$ your problem is the so-called phase retrieval problem (which was motivated by the problem of recovering a function (up to global …

I don't know if this is what you want, but here are two differnet ways: In the case $m=n$: Generate a bunch of random permutation matrices $A_i$ and take a random linear combination $\sum_i \alpha_i …